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otez555 [7]
3 years ago
15

4. A 62.0-kg person, standing on the diving board, dives straight down into the water. Just before striking the water, her speed

is 5.50 m/s. At a time of 1.65 s after she enters the water, her speed is reduced to 1.10 m/s. What is the net average force (magnitude and direction) that acts on her when she is in the water?
Physics
1 answer:
Alekssandra [29.7K]3 years ago
7 0

To solve this problem it is necessary to apply the concepts related to the Moment. The moment in terms of the Force and the time can be expressed as

\Delta P = F\Delta t

F = Force

\Delta t = Time

At the same time the moment can be expressed in terms of mass and velocity, mathematically it can be given as

P = m \Delta v

Where

m = Mass

\Delta v = Change in velocity

Our values are given as

\Delta t=1.65s

By equating the two equations we can find the Force,

F\Delta t = m\Delta v

F = \frac{m\Delta v}{\Delta t}

F = \frac{62(1.1-5.5)}{1.65}

Therefore, the net average force will be:

F = - 165N

The negative symbol indicates that the direction of the force is upwards.

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A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.
V125BC [204]

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar,  F = 534 N

Co-efficient of kinetic friction, \mu_k = 0.3

Now,

The force against the kinetic friction is given as:

f = \mu_k N = u_k Mg

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

f = 0.3\times108\times9.8

or

f = 317.52N

Now, the net force on to the metal safe is

F_{Net}= F-f

Substituting the values in the equation we get

 F_{Net}= 534N-317.52N

or

F_{Net}= 216.48

also,

 

F_{Net}= M\timesacceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=\frac{F_{Net}}{M}

or

 acceleration of the safe=\frac{216.48}{108}

or

 

acceleration of the safe=2.00 m/s^2

Hence, the acceleration of the metal safe will be  2.00 m/s²

3 0
2 years ago
If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T
Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0
2 years ago
______ is most resistant to thermal energy transfers. a. copper wire b. aluminum foil c. water d. foam insulation
Reika [66]

Answer:

Water

Explanation:

Because it does not conduct much energy.

8 0
3 years ago
Calculate the density of the following material: 500 kg gold with a volume of 0.026 m³
Gre4nikov [31]

Answer:

If we have large numbers (b is positive) or small numbers (b is negative), then this way ... 1, and V2i = 100 L, n2i = 5 + 2 + 1 = 8 in vessel 2. ... a good working substance in the barometer.

5 0
3 years ago
Read 2 more answers
g Show that the first derivative of the magnitude of the net magnetic field of the coils (dB/dx) vanishes at the mid- point P re
never [62]

Answer:

Hi Carter,

The complete answer along with the explanation is shown below.

I hope it will clear your query

Pls rate me brainliest bro

Explanation:

The magnitude of the magnetic field on the axis of a circular loop, a distance z  from the loop center, is given by Eq.:

B = NμοiR² / 2(R²+Z²)³÷²

where

R is the radius of the loop

N is the number of turns

i is the current.

Both of the loops in the problem have the same radius, the same number of turns,  and carry the same current. The currents are in the same sense, and the fields they  produce are in the same direction in the region between them. We place the origin  at the center of the left-hand loop and let x be the coordinate of a point on the axis  between the loops. To calculate the field of the left-hand loop, we set z = x in the  equation above. The chosen point on the axis is a distance s – x from the center of  the right-hand loop. To calculate the field it produces, we put z = s – x in the  equation above. The total field at the point is therefore

B = NμοiR²/2 [1/ 2(R²+x²)³÷²   + 1/ 2(R²+x²-2sx+s²)³÷²]

Its derivative with respect to x is

dB /dx=  - NμοiR²/2 [3x/ (R²+x²)⁵÷²   + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]

When this is evaluated for x = s/2 (the midpoint between the loops) the result is

dB /dx=  - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷²   - 3(s/2)/(R²+s²/4)⁵÷² ] =0

independent of the value of s.

8 0
3 years ago
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